a) \(\frac{{ - 3}}{8} = \frac{{ - 3.3}}{{8.3}} = \frac{{ - 9}}{{24}}\)

Vì -9 < -5 nên \(\frac{{ - 9}}{{24}} < \frac{{ - 5}}{{24}}\)

Vậy \(\frac{{ - 3}}{8} < \frac{{ - 5}}{{24}}\).

b) Cách 1: \(\frac{{ - 2}}{{ - 5}} = \frac{2}{5}; \frac{3}{{ - 5}} = \frac{-3}{{5}}\)

Vì 2 > -3 nên \(\frac{2}{5} > \frac{-3}{{5}}\)

Vậy \(\frac{{ - 2}}{{ - 5}} > \frac{3}{{ - 5}}\).

Cách 2: \(\frac{{ - 2}}{{ - 5}} = \frac{2}{5} > 0\) mà \(\frac{3}{{ - 5}} < 0\)

\(\Rightarrow\) \(\frac{{ - 2}}{{ - 5}} > \frac{3}{{ - 5}}\).

c) \(\frac{{ - 3}}{{ - 10}} = \frac{3}{{10}} = \frac{{3.2}}{{10.2}} = \frac{6}{{20}}\)

\(\frac{{ - 7}}{{ - 20}} = \frac{7}{{20}}\)

Vì 6 < 7 nên \(\frac{6}{{20}} < \frac{7}{{20}}\) nên \(\frac{{ - 3}}{{ - 10}} < \frac{{ - 7}}{{ - 20}}\).

d) \(\frac{{ - 5}}{4} = \frac{{ - 5.5}}{{4.5}} = \frac{{ - 25}}{{20}}; \frac{{ 23}}{{-20}}=\frac{{-23}}{{20}} \)

Vì -25 < -23 nên \( \frac{{ - 25}}{{20}} < \frac{{-23}}{{20}} \)

Vậy \(\frac{{ - 5}}{4} < \frac{{23}}{{ - 20}}\).

`3/7-2/5`

`=1/35>0`

`=>3/7>2/5`

`b,9>8`

`=>1/9<1/8`

`=>5/9<5/8`

`d,8/7>1`

`7/8<1`

`=>8/7>7/8`

a) Ta có: \(\dfrac{3}{7}=\dfrac{15}{35}\)

\(\dfrac{2}{5}=\dfrac{14}{35}\)

Do đó: \(\dfrac{3}{7}>\dfrac{2}{5}\)

b) Ta có: 9>8

nên \(\dfrac{5}{9}< \dfrac{5}{8}\)

c) Ta có: \(\dfrac{8}{7}>1\)

\(1>\dfrac{7}{8}\)

Do đó: \(\dfrac{8}{7}>\dfrac{7}{8}\)

a) \(\dfrac{2}{5}=\dfrac{4}{10}\)

\(\dfrac{4}{10}>\dfrac{3}{10}\)

b) \(\dfrac{5}{6}=\dfrac{10}{12}\)

\(\dfrac{7}{12}< \dfrac{10}{12}\)

c) \(\dfrac{1}{2}=\dfrac{2}{4}\)

\(\dfrac{3}{4}< \dfrac{2}{4}\)

d) \(\dfrac{8}{3}=\dfrac{56}{21}\)

\(\dfrac{56}{21}>\dfrac{11}{21}\)

a) \(< \)

b) \(>\)

c) \(< \)

d) \(>\)

e) \(< \)

g) \(>\)

h) \(>\)

k) \(>\)

`a)1<3`

`=>1/5<3/5`

`b)21>9`

`=>8/21<8/9`

`c)3/5<5/5=1`

`d)7/5>5/5=1`

a)3/5>1/5 b)8/21<8/9 c)3/5<1 d)7/5>1

a 1/5<3/5

b 8/21<8/9

c 3/5<1

d 7/5>1

1.a) 3/4 > 5/10

   b) 35/25 > 16/14

2.a) 7/5 > 5/7

  b) 14/16 < 24/21

HT nha

( bạn t.i.c.k cho mik nha, mik cảm ơn )

Tyyyyn

Bạn giúp mình câu này nhé so sánh 317/633 và 371/743 

b) \(\frac{5}{9}\)và \(\frac{5}{8}\) :Quy đồng mẫu số : \(\frac{5}{9}\) = \(\frac{5.8}{9.8}\) = \(\frac{40}{72}\) ; \(\frac{5}{8}\) = \(\frac{5.9}{8.9}\) = \(\frac{45}{72}\)

Vì \(\frac{40}{72}\) < \(\frac{45}{72}\) nên \(\frac{5}{9}\) < \(\frac{5}{8}\)

c)\(\frac{8}{7}\) và \(\frac{7}{8}\) :Quy đồng mẫu số: \(\frac{8}{7}\) = \(\frac{8.8}{7.8}\) = \(\frac{64}{56}\) ; \(\frac{7}{8}\) = \(\frac{7.7}{8.7}\) =\(\frac{49}{56}\)

Vì \(\frac{64}{56}\) > \(\frac{49}{56}\) nên \(\frac{8}{7}\) > \(\frac{7}{8}\)

bạn an đông à cái câu A của bạn sai một chút.

CHÚC BẠN HỌC TỐT !

a)\(\frac{3}{7}\) và\(\frac{2}{8}\) :Quy đồng mẫu số : \(\frac{3}{7}\) = \(\frac{3.8}{7.8}\) = \(\frac{24}{56}\) ; \(\frac{2}{8}\) = \(\frac{2.7}{8.7}\) = \(\frac{14}{56}\)

Vì \(\frac{24}{56}\) > \(\frac{14}{56}\) nên \(\frac{3}{7}\) > \(\frac{2}{8}\)

chim to vai lon

a)5/8 < 8/9

b)8/12> 5/9

c)7/12 < 11/18

a)5/8 < 8/9

b)8/12 > 5/9

c)7/12 < 11/18

a) 5/8 < 8/9

b) 8/12 > 5/9

c) 7/12 < 11/18

a: \(\dfrac{-8}{31}=\dfrac{-8\cdot101}{31\cdot101}=\dfrac{-808}{3131}\)

\(\dfrac{-789}{3131}=\dfrac{-789}{3131}\)

b: Thiếu phân số thứ hai rồi bạn

c: \(\dfrac{1}{n}=\dfrac{n+1}{n\left(n+1\right)}\)

\(\dfrac{1}{n+1}=\dfrac{n}{n\left(n+1\right)}\)

1) \(A=\left(\sqrt{7-\sqrt{21}+4\sqrt{5}}\right)^2=7-\sqrt{21}+4\sqrt{5}\)

\(B=\left(\sqrt{5}-1\right)^2=6-2\sqrt{5}\)

\(\Rightarrow A-B=1-\sqrt{21}+6\sqrt{5}=\left(1+\sqrt{180}\right)-\sqrt{21}>0\)

\(\Rightarrow A>B\Rightarrow\sqrt{7-\sqrt{21}+4\sqrt{5}}>\sqrt{5}-1\)

2) \(C=\left(\sqrt{5}+\sqrt{10}+1\right)^2=5+10+1+10\sqrt{2}+2\sqrt{5}+2\sqrt{10}\)

\(=26+10\sqrt{2}+2\sqrt{5}+2\sqrt{10}>26+10>35=\left(\sqrt{35}\right)^2\)

Vậy \(\sqrt{5}+\sqrt{10}+1>\sqrt{35}\)

3) \(\left(\frac{15-2\sqrt{10}}{3}\right)^2=\frac{225-60\sqrt{10}+40}{9}=\frac{265-60\sqrt{10}}{9}=\frac{265}{9}-\frac{20\sqrt{10}}{3}< 15\)

Vậy nên \(\frac{15-2\sqrt{10}}{3}< \sqrt{15}\)